Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 1 Overview and Descriptive Statistics
Page 34 Problem 1 Answer
Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the given observation.
15.0 13.0 18.0 14.5 12.0 11.0 8.9 8.0
We need to determine the values of the sample mean, sample median, and 12.5% trimmed mean, and compare these values.
Now,Considering the data:
15.0 13.0 18.0 14.5 12.0 11.0 8.9 8.0
We need to find the mean: xˉ=1/n∑xi
=1/8{15+13+18+14.5+12+11+8.9+8.0}
=12.55
For sample median:
We need to arrange the data in ascending order as follows:
8.0 15.08.9 18.011.012.013.014.5
As number of observations n=8 is even then, Median be (8/2)th and (8/2+1)th
So, average of the fourth and fifth values of ordered values gives median:
x~=12+13/2=25/2=12.5 hence,12.5% trimmed mean is determined by eliminating smallest 12.5 % and largest 12.5 % in the sample then using remaining numbers we have to obtain mean.
xtr(12.5)ˉ=8.9+11+12+13+14.5+15/6
=74.4/6
=12.4
From the obtained values, Sample median and trimmed mean are bit smaller than sample mean.
Hence,the values of the sample mean, sample median, and 12.5% trimmed mean obtained as:
Mean:xˉ=12.55
Median:xˉ=12.5
Trimmed mean:xˉ=12.4
Sample median and trimmed mean are bit smaller than sample mean.
Page 34 Problem 2 Answer
Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations:
15.0 13.0 18.0 14.5 12.0 11.0 8.9 8.0
We need to find by how much could the smallest sample observation, currently 8.0, be increased without affecting the value of the sample median.
Now,From the given observations:
15.0 13.0 18.0 14.5 12.0 11.0 8.9 8.0
By how much could the smallest sample observation, currently 8.0, be increased without affecting the value of the sample median:
The smallest observation from the data is 8.0
Without changing median then we have to add such that it should be less than 12.
Hence,we can conclude that by adding 4 to the smallest observation without changing the median.
Therefore,by adding 4 to the smallest observation without effecting the value of sample median.
Page 34 Problem 3 Answer
Given, The minimum injection pressure (psi) for injection molding specimens of high amylose corn was determined for eight different specimens (higher pressure corresponds to greater processing difficulty), resulting in the following observations
15.0 13.0 18.0 14.5 12.0 11.0 8.9 8.0
Suppose we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi.
We need to find whether it is necessary to re-express each observation in ksi, or can the values calculated in part (a) be used directly.
Now,The sample mean,median and trimmed mean of the observations obtained in 35(a) as follows:
Mean: xˉ=12.55 Median: xˉ=12.5 Trimmed mean: xˉ=12.4
To convert them form pounds per square inch to kilogram per square inch:
We need to divide the obtained values by 2.2 such that it gets converted to ksi.
New sample mean
=12.55/2.2
=5.70
New sample median
=12.50/2.2
=5.68
New Trimmed mean
=12.4/2.2
=5.64
Hence,it is verified that there is no need to recalculate the each value to convert them form psi to ksi.
So,as we want the values of the sample mean and median when the observations are expressed in kilograms per square inch (ksi) rather than psi then there is no need to recalculate the each value.
We can obtain them by directly dividing each obtained values with 2.2.
New sample mean=5.70 ksi
New sample median=5.68 ksi
Trimmed mean=5.64 ksi
Page 34 Problem 4 Answer
Given,Using the article “Snow Cover and Temperature Relationships in North America and Eurasia” the statistical techniques to relate the amount of snow cover on each continent to average continental temperature:
There are ten observations on October snow cover for Eurasia during a period.
We need to find report a representative, or typical, value of October snow cover for this period.
Given data- 6.5/7.912.0/21.914.9/12.510.0/14.510.7/9.2
For the given data, 21.9 is an outlier, so trimmed mean would be good choice for the researcher.
For the trimmed mean remove the smallest and the largest values-
xtr(10)=12.0+14.9+10.0+10.7+7.9+12.5+14.5+9.2/8
=91.7/8
=11.4625
Hence, based on the article “Snow Cover and Temperature Relationships in North America and Eurasia” using statistical techniques to relate the amount of snow cover on each continent to average continental temperature report as representative, or typical, value of October snow cover for the period is represented by 10% trimmed mean which is 11.4625.
Page 40 Problem 5 Answer
Given,Based on the study of Life distribution of micro-drills listed in increasing order on drill life time:
Using the Data on drill lifetime in certain brass alloy:
We need to find sample median,25% trimmed mean, 10% trimmed mean, and sample mean, and compare them.
The ordered values are-11,14,20,23,31,36,39,44,47,50,59,61,65,67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,112,118,123,136,139,141,148,158,161,168,184,206,248,263,289,322,388,513
The given data has even values so the sample median is-
Therefore, Median =x25+x26/2
=91+93/2
=184/2
=92
For 25% trimmed mean, the trimmed value is,
Trimmed value =n× Percentage
=50×0.25
=12.5≈13
So delete 13 Now, the remaining data is, observations on both sides-
67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,105,112,118,123,136,139.
Therefore, the mean is,
xTr−(0.25)
=1/nTr(∑xTr)
=1/24(67+68+71+……+139)
=1/24(2274)
=94.75
For 10% trimmed mean, the trimmed value is,
Trimmed value =n× percentage
=50×0.10
=5
So delete 5 observations on both sides-
Now, the remaining data is,11,14,20,23,31,36,39,44,47,50,59,61,65,67,68,71,74,76,78,79,81,84,85,89,91,93,96,99,101,104,105,112,118,123,136,139,141,148,158,161,168,184,206,248,263,289,322,388,513
Therefore, the mean is,
xTr−(0.10)
=1/nTr(∑xrr)
=1/40(36+39+44+47+…+248)
=1/40(4089)
=102.225
The sample mean.
xˉ=1/n∑xi
=1/50(11+14+20+….+388+513)
=1/50(5963)
=119.26
So, we can observe that the sample mean has more value than the 10% trimmed mean and 10% trimmed mean has more value than the 25% trimmed mean.
Thus, we can conclude that the mean value is decreasing when trimming the data.
Hence,the sample median is92,25% trimmed mean is 94.75,10% trimmed mean is 102.225 and sample mean is 119.26.
On comparing the obtained values, we can conclude that the mean value is decreasing when trimming the data.
Page 41 Problem 6 Answer
Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows:
A sample of n=10 automobiles for a crash test.
We need to find-the value of the sample proportion of successes.
SSFFSSSFSS
S means success and F damage.
So here we have 7 successes and 3 damages , total 10 tests.
The sample proportion of successes x/n
=7/10
=0.7
Hence,the value of the sample of automobiles selected, and each was subjected to a 5 -mph crash test and proportion of successes is 0.7
Page 41 Problem 7 Answer
Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows:
A sample of n=10 automobiles for a crash test.
We need to find xˉ and compare it with x/n.
Here we have to replace each S with 1 a and each F with a 0 .
So it becomes 1,1,0,0,1,1,1,0,1,1
Herexˉ=7/10
=0.7
So it can be observed that xˉ is similar to the proportion of success xn.
Hence,after replacing each S with a 1 and each F with a 0 ,the mean:
xˉ=0.7 and it is equal to xn.
Page 41 Problem 8 Answer
Given, Denoting a car with no visible damage by S (for success) and a car with such damage by F, results were as follows
A sample of n=25 automobiles for a crash test.
We need to find the number of these would have to be S′s to give xn=.80 .
We currently have actually 7 S′s and 3F′s-And we want xn=0.8n would be 25,
x=25∗0.8
=20
So we need 20 S′s and we have 7.
So 13S′s from the new 15 cars is required in order to satisfy the condition.
So, by including 15 more cars in the experiment 13S′s
would have to be from new 15 tests, so that xn=0.8